Solutions to Homework 7

Transcription

1 Solutions to Homework 7 Exercise #3 in section 5.2: A rectangular box is inscribed in a hemisphere of radius r. Find the dimensions of the box of maximum volume. Solution: The base of the rectangular box lies in the plane that contains the base of the hemisphere. If the center of the hemisphere lies at the origin, then in coordinates the hemisphere has equation z = r 2 x 2 y 2. The vertices of the inscribed rectangular box will, after rotating if necessary, lie at points of the form (±a, ±b c) or (±a, ±b, 0). The volume of this box will be (2a)(2b)c = 4abc, where a, b > 0. Moreover, since the vertex (a, b, c) lies on the hemisphere, c = r 2 a 2 b 2. And therfore, the volume of the box can be expressed as a fucntion of a and b alone as follows: V (a, b) = 4ab r 2 a 2 b 2. The goal is to determine the maximum value of this function. The values of a and b satisfy 0 < a 2 + b 2 < r. The function V is continuous on the region 0 a 2 + b 2 r, and this region is compact. Therefore, V attains a maximum value. The maximum value must lie in the interior of this region since V = 0 if either 0 = a 2 + b 2 or a 2 + b 2 = r. The critical points of V are determined by solving DV (a, b) = 0 or, equivalently, solving V a = 0 and V b = 0. Using the product rule, one obtains the following: V a = 4b r 2 a 2 b 2 4a 2 b r 2 a 2 b 2. The formula for V b is the same except that the roles of a and b are interchanged. The equation V a = 0 can be rewritten as follows after cancelling b (which is non-zero), moving the negative expression to the other side of the equation, and cross multiplying and regrouping: r 2 = 2a 2 + b 2 Similarly, V b = 0 can be rewritten as r 2 = 2b 2 + a 2. Multiplying this equation by two and subtracting the equation above results in r 2 = 3b 2 = a = b = r 3.

2 Since this is the only critical point in the interior of the region 0 a 2 + b 2 r 2 and since V attains a maximum in the interior and since V is differentiable in the interior, the maximum value of V must occur at a = b = r/ 3. Therefore, the maximum volume of such a box is 4r 3 /3 3. Notes: This problem can be solved using the methods of calculus of a single variable. It is clear from symmetry that the base of the box must be a square. If this square has an edge of length b and if the box has a height of h, then V = b 2 h. Let a be the length of the segment from the center of the square to a vertex. Then a 2 + h 2 = r 2 and b = 2a. Therefore, V = 2(r 2 h h 3. Since V = 0 when h = r/ 3 and V < 0, the V has a maximum value of 4r 3 /3 3. Exercise #4 in section 5.2: Determine the maximum and minimum values of the function f(x, y) = x 2 + 2y 2 2x on the disk D = {(x, y) : x 2 + y 2 2. Solution: The answer is in the textbook. You will find that there is only one crictial point in the interior of the region D, namely (1, 0); and one computes that f(1, 0) = 1. On the boundary of the disk D, one uses the fact that x 2 + y 2 = 2 to study f restricted to the boundary: f = x 2 + 2(2 x 2 ) 2x = x 2 2x + 4. The above is valid only if 2 x 2. Since f = 2x 2 and f = 0 when x = 1, the extreme values of f restricted to the boundary of D must belong to the following list f( 2) = , f( 1) = 5, f( 2 = Among all of the possible extreme values (including f(1, 0) = 1), one sees that 1 is the minimum value and that 5 is the maximum value. Notes: This is a very good example of how the extreme values of a function can occur either in the interior of a compact region or on the boundary of a compact region. To analayze a function on the boundary of a region, one uses the additional constraint and solves the new constrained optimization problem. Exercise #1 (b, d, e, h) in section 5.3: Classify the critical points of each function below.

3 Solution: (b): f(x, y) = xy + x y. Since f x = y + 1 and f y = x 1, the only critical point is (1, 1). The second order partial derivatives are f xx = 0, f yx = 1, f xy = 1, and f yy = 0. Therefore, the Hessian matrix defines an indefinite quadratic form (the determinant of the Hessian matrix is 1 and Corollary 3.4 applies). Hence, f has a saddle point at (1, 1). (d): f(x, y) = x 2 3x 2 y + y 3. f x = 2x 6xy, f y = 3x 2 + 3y 2. f y = 0 implies that x = ±y. If x = y, then f x = 0 implies that 2x(1 3x) = 0 and so (0, 0) and (1/3, 1/3) are critical points. If x = y, then f x = 0 implies that 2x(1 + 3x) = 0 and so ( 1/3, 1/3) is also a critical point. The second order partial derivatives are f xx = 2 3y, f yx = 6x, f xy = 6x, f yy = 6y. Let H(x, y) = (2 3y)6y 36x 2 be the determinant of the Hessian matrix at (x, y). Since H(0, 0) = 0, Corollary 3.4 does not apply. However, one can deduce that (0, 0) is a saddle point directly from the defintion: every open ball containing (0, 0) contains both points such that f is positive and points such that f is negative. Given any open set U containing (0, 0), choose n N such that the ball of radius 1/n with center (0, 0) is contained in U. Since f(1/2n, 0) > 0 and f(1/2n, 1/4n) < 0 (check this). On the othe other hand, Corollary 3.4 can be applied to the other critical points: H(1/3, 1/3) < 0, so there is a saddle point at (1/3, 1/3); and H( 1/3, 1/3) < 0, so there is also a saddle point at ( 1/3, 1/3). (e) f(x, y) = x 2 y + x 3 x 2 + y 2. The solution is similar to the previous two problems. The critical points are (0, 0), (1, 1/2), and (2, 2). These critical points are, respectively, a saddle point, a local minimum, and a saddle point. (h) f(x, y) = x 2 y 4xy y 2. The solution is similar to the previous three problems. The critical points are (0, 0), (4, 0) and (2, 2). These are, respectively, a saddle point, a saddle point, and a local maximum. Exercise # 1 in section 5.4: Solution: (a) Find the minimum value of f(x, y) = x 2 + y 2 on the curve x + y = 2. Why is there no maximum?

4 There is no maximum value since on this curve, f = x 2 + (2 x) 2 = 2x 2 4x + 4, which is a function which increases without bound. To find the minimum, one can use the method of Lagrange multipliers. (The methods of calculus of a single variable are also appropriate.) Let g = x + y 2. Then f = [2x, 2y] T and g = [1, 1] T. Since g is non-zero it has rank one at every point of g 1 (0) and so Theorem 4.1 applies. Let λ be the Lagrange multiplier. One solves the system 2x = λ 1, 2y = λ 1, x + y 2 = 0. There is a unique solution: x = y = 1 and λ = 2. The value of f(1, 1) is equal to 2. This is the minimum value of f since f has a minimum value when restricted to the compact set g 1 (0) {0 x 2} and since f(0, 2) and f(2, 0) are greater than f(1, 1). (Theorem 4.1 states that since f is differentiable and if f has an extreme value at a point a on g 1 (0), then a must be a solution to the system of equations above. What is written above is a formal argument that, in fact, the solution to this system is a global minimum value of f (as opposed to a local extremum which is not a global extremum). (b) Find the maximum value of g(x, y) = x + y on the curve x 2 + y 2 = 2. Is there a minimum? There is a maximum and a minimum value since the set {x 2 + y 2 = 2} is compact and g is continuous. One can determine these global extrema by the method of Lagrange multipliers (or by methods of calculus of a single variable). Let h = x 2 + y 2 2. g = [1, 1] T and h = [2x, 2y] T. Since x = y = 0 is not a point in h 1 (0), Theorem 4.1 applies. (The rank of Dh is equal to one at every point of h 1 (0).) One solve the system 1 = λ(2x), 1 = λ(2y), x 2 + y 2 2 = 0. The first two equations implie that λ 0, x 0, and y 0. Divide the first equation by the second to obtain x = y. By the third equation, x = y = 1 or x = y = 1. (It is not possible for x and y to have opposite signs by the first two equations.) Therefore, the extreme values of g must occur at one of the following points: (1, 1), or ( 1, 1). Evaluating g at each of these points, one determines that g has a maximum value of 2 and a minimum value of 2. (c) How are the questions (and answers) in parts (a) and (b) related?

5 This is an open ended question. Here is my answer. Both problems concern the problem of finding points which are closest or farthest from a given curve. For part (a) this is clear: the problem asks one to determine the closest point on the line x + y = 2 to the origin. (There is no farthest point on the line to the origin.) For part (b) this is less clear. However, one might interpret g(x, y) = x + y as measuring the distance to the origin in a certain sense, at least if x and y are positive. And in this case, we are asked to determine the closest and farthest points on the circle x 2 + y 2 = 2 to the origin. My answer is guided (or misguided?) by my previous encounters with what are know as the l 2 and l 1 norms. You can find additional information about this topics on Wikipedia, for instance. Exercise # 2 in section 5.4 Solution: The answer to this problem is in the textbook. There is a maximum and a minimum value since the curve x 2 + y 2 2y = 0 is a compact set; it is a circle centered at (0, 1). You should solve this problem using the method of Lagrange multipliers (to practice this method), and then check your work by also computing a solution using methods of calculus of a single variable.